Question

9. A skateboarder of mass m is standing on the edge of a drainage channel, as shown. The left side, where the skateboarder starts, is elevated at an angle θ; the right side is elevated at an angle φ. The slopes on either side are smooth, and the skateboard moves over them with essentially no friction, but the flat bottom of width b is covered with a little sand, and the skateboard experiences a small amount of rolling friction there, with μk known.

The skateboarder starts a distance d up the left-hand side. They roll down the left side, across the sand-filled bottom, and up the right side.

(a) Determine the maximum distance d2 that the skateboarder makes it up the right side. (This is the diagonal distance, not the height.)

(b) After rolling up and back down the right side, the skateboarder will come back to the left side. How far will they travel back up the left side?

(c) Suppose that you know numeric values as follows: m=75kg, θ = 30◦  φ = 40◦ μr = 0.05 d=4m b=7m

How many times will the skateboarder travel across the sandy bottom of the channel before coming to rest? Explain the approach behind your solution fully.

Friction here; width b - - - - - - - - - - - ) С.

Answer 1

Solution la conservation of energy. mgl d sino) - Me mg b = mg da sin e asino - ble dz sinq da = dsin o-b sing (6.) conservation of energy. mg da sing - Hi mg b = mg de sine. de = dz sinu-ble sino dz = (asino - blk) sing- blk since da = sino d sino -abuk sino (c.) In ist trip dz = - In and trip , 'da = d sin & -ablk sino d sino – 4 blk sino - In nth trip, V - dnth d sino-ano hk Sino dsino - anblk =0 For, doon =0 » n= d sino 26 Mk

n= (4m )( sin 30°) 2 *(7m)10.05) - 2.857 thus, 2.857 trips that is more than 2.5 trips and less than 3 trips. I trip E a time on sandy bottom 2,5 trips = 5 times Hence, 5 times the skateboarder will cross the sandy bottom

The skateboarder will cross the sandy bottom completely 5 times.